Question

If $\tan \theta =\frac{1}{\sqrt{3}}$ then evaluate $\frac{\cos e{c}^2\theta -{\sec }^2\theta }{\cos e{c}^2\theta +{\sec }^2\theta }$

Answer 1

$tan\theta =\frac{1}{\sqrt{3}}\Rightarrow tan\theta =tan{30}^{\circ }\Rightarrow \theta ={30}^{\circ }$

$\frac{cose{c}^2\theta -se{c}^2\theta }{cose{c}^2\theta +se{c}^2\theta }=\frac{cose{c}^2{30}^{\circ }-se{c}^2{30}^{\circ }}{cose{c}^2{30}^{\circ }+se{c}^2{30}^{\circ }}$

$=\frac{2^2-(2/\sqrt{3}{)}^2}{2^2+(2/\sqrt{3}{)}^2}$

$=\frac{4-4/3}{4+4/3}=\frac{12-4}{12+4}=\frac{8}{16}=\frac{1}{2}$